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2 years ago

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An asset is purchased on April 1st and has an annual depreciation amount of $3,600. Using the straight-line method of depreciati

on, what is the depreciation expense on December 31st?
$3,000

$2,700

$2,900

$3,240

2 answers:

Free_Kalibri [48]2 years ago

5 0

From April to December =9 months

the depreciation expense
3,600×(9÷12)=2,700

Vlad1618 [11]2 years ago

4 0

The answer to your question is 2,700

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The mean height of women in a country (ages 20-29) is 64 4 inches A random sample of 50 women in this age group is selected What

Simora [160]

Answer:

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 64.4 inches, standard deviation of 2.91

This means that $\mu = 64.4, \sigma = 2.91$

Sample of 50 women

This means that $n = 50, s = \frac{2.91}{\sqrt{50}}$

What is the probability that the mean height for the sample is greater than 65 inches?

This is 1 subtracted by the p-value of Z when X = 65. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{65 - 64.4}{\frac{2.91}{\sqrt{50}}}$

$Z = 1.46$

$Z = 1.46$ has a p-value of 0.9279

1 - 0.9279 = 0.0721

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

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The produce manager at a market orders 10 lb of tomatoes, 20 lb of zucchini, and 40 lb of onions from a local farmer one week.

rewona [7]

Answer:

(A) $A=\left[\begin{array}{ccc}10&20&40\end{array}\right]$

(B) $B=\left[\begin{array}{ccc}11&22&44\end{array}\right]$

(C) $A+B=\left[\begin{array}{ccc}21&42&84\end{array}\right]$

Step-by-step explanation:

The manager ordered 10 lb of tomatoes, 20 lb of zucchini, and 40 lb of onions from a local farmer one week.

(A)

Matrix <em>A</em> represents the amount of each item ordered. It is 1 × 3 matrix.

Then matrix <em>A</em> is:

$A=\left[\begin{array}{ccc}10&20&40\end{array}\right]$

(B)

Next week the manager increases the order of all the products by 10%.

Then the amount of new orders are:

Tomatoes $=10\times [1+\frac{10}{100}]=10\times1.10=11$

Zucchini $=20\times [1+\frac{10}{100}]=20\times1.10=22$

Onions $=40\times [1+\frac{10}{100}]=40\times1.10=44$

Th matrix <em>B</em> represents the amount of each order for the next week. Then matrix <em>B</em> is:

$B=\left[\begin{array}{ccc}11&22&44\end{array}\right]$

(C)

Add the two matrix <em>A</em> and <em>B</em> as follows:

$A+B=\left[\begin{array}{ccc}10&20&40\end{array}\right]+\left[\begin{array}{ccc}11&22&44\end{array}\right]\\=\left[\begin{array}{ccc}(10+11)&(20+22)&(40+44)\end{array}\right]\\=\left[\begin{array}{ccc}21&42&84\end{array}\right]$

The entries of the matrix (<em>A</em> + <em>B</em>) represent the amount of tomatoes, zucchini and onions ordered for two weeks.

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2 years ago